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P [Preliminaries] - INSTRUCTOR'S SOLUTIONS MANUAL SECTION...

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION P.1 (PAGE 10) CHAPTER P. PRELIMINARIES Section P.1 Real Numbers and the Real Line (page 10) 1. 2 9 = 0 . 22222222 · · · = 0 . 2 2. 1 11 = 0 . 09090909 · · · = 0 . 09 3. If x = 0 . 121212 · · · , then 100 x = 12 . 121212 · · · = 12 + x . Thus 99 x = 12 and x = 12 / 99 = 4 / 33. 4. If x = 3 . 277777 · · · , then 10 x 32 = 0 . 77777 · · · and 100 x 320 = 7 + ( 10 x 32 ) , or 90 x = 295. Thus x = 295 / 90 = 59 / 18. 5. 1 / 7 = 0 . 142857142857 · · · = 0 . 142857 2 / 7 = 0 . 285714285714 · · · = 0 . 285714 3 / 7 = 0 . 428571428571 · · · = 0 . 428571 4 / 7 = 0 . 571428571428 · · · = 0 . 571428 note the same cyclic order of the repeating digits 5 / 7 = 0 . 714285714285 · · · = 0 . 714285 6 / 7 = 0 . 857142857142 · · · = 0 . 857142 6. Two different decimal expansions can represent the same number. For instance, both 0 . 999999 · · · = 0 . 9 and 1 . 000000 · · · = 1 . 0 represent the number 1. 7. x 0 and x 5 define the interval [0 , 5]. 8. x < 2 and x ≥ − 3 define the interval [ 3 , 2 ) . 9. x > 5 or x < 6 defines the union ( −∞ , 6 ) ( 5 , ) . 10. x ≤ − 1 defines the interval ( −∞ , 1]. 11. x > 2 defines the interval ( 2 , ) . 12. x < 4 or x 2 defines the interval ( −∞ , ) , that is, the whole real line. 13. If 2 x > 4, then x < 2. Solution: ( −∞ , 2 ) 14. If 3 x + 5 8, then 3 x 8 5 3 and x 1. Solution: ( −∞ , 1] 15. If 5 x 3 7 3 x , then 8 x 10 and x 5 / 4. Solution: ( −∞ , 5 / 4] 16. If 6 x 4 3 x 4 2 , then 6 x 6 x 8. Thus 14 7 x and x 2. Solution: ( −∞ , 2] 17. If 3 ( 2 x ) < 2 ( 3 + x ) , then 0 < 5 x and x > 0. Solution: ( 0 , ) 18. If x 2 < 9, then | x | < 3 and 3 < x < 3. Solution: ( 3 , 3 ) 19. Given: 1 /( 2 x ) < 3. CASE I. If x < 2, then 1 < 3 ( 2 x ) = 6 3 x , so 3 x < 5 and x < 5 / 3. This case has solutions x < 5 / 3. CASE II. If x > 2, then 1 > 3 ( 2 x ) = 6 3 x , so 3 x > 5 and x > 5 / 3. This case has solutions x > 2. Solution: ( −∞ , 5 / 3 ) ( 2 , ) . 20. Given: ( x + 1 )/ x 2. CASE I. If x > 0, then x + 1 2 x , so x 1. CASE II. If x < 0, then x + 1 2 x , so x 1. (not possible) Solution: ( 0 , 1]. 21. Given: x 2 2 x 0. Then x ( x 2 ) 0. This is only possible if x 0 and x 2. Solution: [0 , 2]. 22. Given 6 x 2 5 x ≤ − 1, then ( 2 x 1 )( 3 x 1 ) 0, so either x 1 / 2 and x 1 / 3, or x 1 / 3 and x 1 / 2. The latter combination is not possible. The solution set is [1 / 3 , 1 / 2]. 23. Given x 3 > 4 x , we have x ( x 2 4 ) > 0. This is possible if x < 0 and x 2 < 4, or if x > 0 and x 2 > 4. The possibilities are, therefore, 2 < x < 0 or 2 < x < . Solution: ( 2 , 0 ) ( 2 , ) . 24. Given x 2 x 2, then x 2 x 2 0 so ( x 2 )( x + 1 ) 0. This is possible if x 2 and x ≥ − 1 or if x 2 and x ≤ − 1. The latter situation is not possible. The solution set is [ 1 , 2]. 25. Given: x 2 1 + 4 x . CASE I. If x > 0, then x 2 2 x + 8, so that x 2 2 x 8 0, or ( x 4 )( x + 2 ) 0. This is possible for x > 0 only if x 4. CASE II. If x < 0, then we must have ( x 4 )( x + 2 ) 0, which is possible for x < 0 only if x ≥ − 2. Solution: [ 2 , 0 ) [4 , ) . 26. Given: 3 x 1 < 2 x + 1 . CASE I. If x > 1 then ( x 1 )( x + 1 ) > 0, so that 3 ( x + 1 ) < 2 ( x 1 ) . Thus x < 5. There are no solutions in this case.
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